\(=\sqrt{\dfrac{b+1}{b^2}}=\left[{}\begin{matrix}\dfrac{\sqrt{b+1}}{b}\left(b>0\right)\\-\dfrac{\sqrt{b+1}}{b}\left(-1\le b< 0\right)\end{matrix}\right.\)

Giúp mình với 

có nghĩa khi và
Nếu thì
Nếu thì Tương tự như vậy ta có:
Nếu thì
Nếu thì Ta có:
Điều kiện để căn thức có nghĩa là hay Do đó:
Nếu b>0 thì
Nếu thì Điều kiện để có nghĩa là hay
Cách 1.
=
Cách 2. Biến mẫu thành một bình phương rồi áp dụng quy tắc khai phương một thương: Điều kiện để có nghĩa là hay xy>0.
Do đó

(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)

#Học tốt!!!

\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)

\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)

\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)

\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)

\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)

\(a\cdot b\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{a\cdot b}\)

\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)

\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)

\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)

\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)

tuổi con HN là :

50 : ( 1 + 4 ) = 10 ( tuổi )

tuổi bố HN là :

50 - 10 = 40 ( tuổi )

hiệu của hai bố con ko thay đổi nên hiệu vẫn là 30 tuổi

ta có sơ đồ : bố : |----|----|----|

                  con : |----| hiệu 30 tuổi

tuổi con khi đó là :

 30 : ( 3 - 1 ) = 15 ( tuổi )

số năm mà bố gấp 3 tuổi con là :

 15 - 10 = 5 ( năm )

       ĐS : 5 năm

mình nha

mình đâu có hỏi tuổi của ai??????????????????

Giải:

a) \(\dfrac{a}{b}\sqrt{\dfrac{b}{a}}\)

\(=\sqrt{\dfrac{b}{a}.\left(\dfrac{a}{b}\right)^2}\)

\(=\sqrt{\dfrac{b}{a}.\dfrac{a^2}{b^2}}\)

\(=\sqrt{\dfrac{a^2.b}{ab^2}}\)

\(=\sqrt{\dfrac{a}{b}}\)

Vậy ...

b) \(3xy\sqrt{\dfrac{2}{xy}}\)

\(=\sqrt{\dfrac{2.\left(3xy\right)^2}{xy}}\)

\(=\sqrt{\dfrac{2.9x^2y^2}{xy}}\)

\(=\sqrt{18xy}\)

Vậy ...

\(\sqrt{\dfrac{1}{600}}\)=\(\sqrt{\dfrac{1}{10^2\cdot6}}\)=\(\sqrt{\dfrac{1\cdot6}{10^2\cdot6\cdot6}}\)=\(\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}\)=\(\sqrt{\dfrac{11\cdot540}{540\cdot540}}\)=\(\dfrac{\sqrt{5940}}{540}\)=\(\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}\)=\(\sqrt{\dfrac{3\cdot50}{50\cdot50}}\)=\(\dfrac{\sqrt{150}}{50}\)=\(\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}\)=\(\sqrt{\dfrac{5\cdot98}{98\cdot98}}=\dfrac{\sqrt{490}}{98}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)

\(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{6}}{60}\)

\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)

\(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{6}}{10}\)

\(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{10}}{14}\)

\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)

ĐS:

+ Ta có:

2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)

                   =2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5

                   =2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).

+ Ta có:

3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)

                    =3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7

                    =3(√10−√7)3=√10−√7=3(10−7)3=10−7.

+ Ta có:

1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)

=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y

+ Ta có:

2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)

=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.

\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)

\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)

\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)

\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)

33+1=3(3−1)(3+1)(3−1)=3(3−1)3−1=3(3−1)2.

23−1=2(3+1)(3−1)(3+1)=2(3+1)3−1=3+

2+32−3=(2+3)(2+3)(2−3)(2

b3+b=b(3−b)(3+b)(3−b)=b(3−b)32−(b)2=b(3−b)9−b.

p2.p−1=p(2.p+1)(2.p−1)(2.p+

33+1=3(3&#x2212;1)(3+1)(3&#x2212;1)=33&#x2212;3.1(3)2&#x2212;12" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">33+1=3(3−1)(3+1)(3−1)=33−3.1(3)2−12

33&#x2212;33&#x2212;1=33&#x2212;32" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">33−33−1=33−32.

23&#x2212;1=2(3+1)(3&#x2212;1)(3+1)=2(3+1)(3)2&#x2212;12" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">23−1=2(3+1)(3−1)(3+1)=2(3+1)(3)2−12

2(3+1)3&#x2212;1=2(3+1)2=3+1" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">2(3+1)3−1=2(3+1)2=3+1.

2+32&#x2212;3=(2+3).(2+3)(2&#x2212;3)(2+3)=(2+3)222&#x2212;(3)2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">2+32−3=(2+3).(2+3)(2−3)(2+3)=(2+3)222−(3)2

22+2.2.3+(3)24&#x2212;3" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">22+2.2.3+(3)24−3

7+431=7+43" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">7+431=7+43.

b3+b=b(3&#x2212;b)(3+b)(3&#x2212;b)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">b3+b=b(3−b)(3+b)(3−b)

b(3&#x2212;b)32&#x2212;(b)2=b(3&#x2212;b)9&#x2212;b;(b&#x2260;9)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">b(3−b)32−(b)2=b(3−b)9−b;(b≠9).

p2p&#x2212;1=p(2p+1)(2p&#x2212;1)(2p+1)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">p2p−1=p(2p+1)(2p−1)(2p+1)

p(2p+1)(2p)2&#x2212;12=p(2p+1)4p&#x2212;1" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:19.36px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">p(2p+1)(2p)2−12=p(2p+1)4p−1

#Ye Chi-Lien

\(\frac{3}{\sqrt{3}+1}=\frac{3\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3\sqrt{3}-3}{3-1}=\frac{3\sqrt{3}-3}{2}\)

\(\frac{2}{\sqrt{3}-1}=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{2\left(\sqrt{3}+1\right)}{3-1}=\sqrt{3}-1\)

\(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3}=\left(2+\sqrt{3}\right)^2=4+4\sqrt{3}+3=7+4\sqrt{3}\)

\(\frac{b}{3+\sqrt{b}}=\frac{b\left(3-\sqrt{b}\right)}{\left(3+\sqrt{b}\right)\left(3-\sqrt{b}\right)}=\frac{b\left(3-\sqrt{b}\right)}{9-b}\)

\(\frac{p}{2\sqrt{p}-1}=\frac{p\left(2\sqrt{p}+1\right)}{\left(2\sqrt{p}-1\right)\left(2\sqrt{b}+1\right)}=\frac{p\left(2\sqrt{b}+1\right)}{4p-1}\)

+) \dfrac{3}{\sqrt{3}+1}=\dfrac{3(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\dfrac{3(\sqrt{3}-1)}{3-1}=\dfrac{3(\sqrt{3}-1)}{2}3​+13​=(3​+1)(3​−1)3(3​−1)​=3−13(3​−1)​=23(3​−1)​.

+) \dfrac{2}{\sqrt{3}-1}=\dfrac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\dfrac{2(\sqrt{3}+1)}{3-1}=\sqrt{3}+13​−12​=(3​−1)(3​+1)2(3​+1)​=3−12(3​+1)​=3​+1.

+) \dfrac{2+\sqrt{3}}{2-\sqrt{3}}=\dfrac{(2+\sqrt{3})(2+\sqrt{3)}}{(2-\sqrt{3})(2+\sqrt{3})}=\dfrac{4+4 \sqrt{3}+3}{2^{2}-(\sqrt{3})^{2}}=7+4 \sqrt{3}2−3​2+3​​=(2−3​)(2+3​)(2+3​)(2+3)​​=22−(3​)24+43​+3​=7+43​.

+) \dfrac{b}{3+\sqrt{b}}=\dfrac{b(3-\sqrt{b})}{(3+\sqrt{b})(3-\sqrt{b})}=\dfrac{b(3-\sqrt{b})}{3^{2}-(\sqrt{b})^{2}}=\dfrac{b(3-\sqrt{b})}{9-b}3+b​b​=(3+b​)(3−b​)b(3−b​)​=32−(b